# moments of area/inertia

• posted

Hello,

I am working on my aerospace engineering dissertation for active damping of wind turbine vibration. The problem I have here is to do with strucures/mechanics.

I am trying to find the centroid of an arbitrary aerofoil section, but it has been at least a year since I found the centroid of anything, and I am slighlty confused. The method I used gave a result that looks wrong

I have a polynomial expression that describes the shape of the section. Let's just say that y=f(x).

I know to find the centroid with respect to x you use the double integral: Qx=int(y)dA=int(y)dxdy. but

I can easily integrate w.r.t. x, but because y is a funtion of x, and y is not constant like a rectangle would be, I don't really know what to do.

I tried multiplying the integral with respect to x and then multiplying by the maximum thickness (in the y-direction) of the aerofoil. This gave a result that ooks far too big to be accurate.

I could use the trapezium rule, adding rectangles and triangles to approximate the profile, but do not really want to loose any accuracy if at all possible.

I'm not really sure how to define dy. Is it just the y-ordinate at a given x? Something tells me that at a given x, y>>dy.

If anybody can point me in the right direction or tell me

• posted

Forget going for a closed form solution BTDT ...years ago I toyed with writing a mass moment of inertia program for bodies of revolution (BASIC then Fortran) closed form but settled on a "trapezoidal method" numerical solution

I suggest you go for a numerical solution, it will be close enough & you get there quicker

cheers Bob

• posted

Here's a possible method if the slope of your aerofoil never exceeds, say, 1.0. Create the following four functions defining the geometry of your aerofoil cross section.

y1(x) = aerofoil lower skin midsurface, y2(x) = aerofoil upper skin midsurface, t1(x) = skin thickness (perpendicular to skin tangent) of lower skin, t2(x) = skin thickness (perpendicular to skin tangent) of upper skin.

Then,

A = integral{[t1(x)/cos(theta1(x)) + t2(x)/cos(theta2(x))]*dx,0,x2}, yc = integral{[y1(x)*t1(x)/cos(theta1(x)) + y2(x)*t2(x)/cos(theta2(x))]*dx,0,x2}/A, Ix = integral{[((y1(x)-yc)^2)*t1(x)/cos(theta1(x)) + ((y2(x)-yc)^2)*t2(x)/cos(theta2(x))]*dx,0,x2},

where theta1(x) = atan(dy1(x)/dx), theta2(x) = atan(dy2(x)/dx), A = aerofoil cross-sectional area, yc = centroid y coordinate, Ix = area moment of inertia (second moment of area) about line passing through centroid parallel to x axis, and dy1(x)/dx = derivative of y1(x) w.r.t. x. The integration limits are x = 0 to x2. Similarly,

xc = integral{[x*t1(x)/cos(theta1(x)) + x*t2(x)/cos(theta2(x))]*dx,0,x2}/A, Iy = integral{[((x-xc)^2)*t1(x)/cos(theta1(x)) + ((x-xc)^2)*t2(x)/cos(theta2(x))]*dx,0,x2}.

Similarly,

Ixy = integral{[((x-xc)(y1(x)-yc))*t1(x)/cos(theta1(x)) + ((x-xc)(y2(x)-yc))*t2(x)/cos(theta2(x))]*dx,0,x2},

where Ixy = product of inertia. Note that dx/cos(theta1(x)) = differential arc length = ds. Make sure your slope in y1(x) or y2(x) doesn't ever exceed, say, 1.0, because the above solution could then become inaccurate. If the slope ever exceeds, say, 1.0, you would need to add those steep segments to your integrals manually, instead of integrating them.

• posted

An attempt could be worthwhile,but not too sure,as we know how to use Green's Theorem to compute areas of closed loops by an elegant formula.Similarly the first integral affords a simplification and hence also the centroid.

• posted

An attempt could be worthwhile,but not too sure,as we know how to use Green's Theorem to compute areas of closed loops by an elegant formula.Similarly the first integral affords a simplification and hence also the centroid.

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